Problem: [Æsir - CHAOS](https://soundcloud.com/kivawu/aesir-chaos)

[Æsir - V.](https://soundcloud.com/kivawu/aesir-v)

"Everything has been planned out. No more hidden concerns. The condition of Cytus is also perfect.

The time right now...... 00:01:12......

It's time."

The emotion samples are now sufficient. After almost 3 years, it's time for Ivy to awake her bonded sister, Vanessa.

The system inside A.R.C.'s Library core can be considered as an undirected graph with infinite number of processing nodes, numbered with all positive integers (1, 2, 3, …). The node with a number x (x > 1), is directly connected with a node with number (x)/(f(x)), with f(x) being the lowest prime divisor of x.

Vanessa's mind is divided into n fragments. Due to more than 500 years of coma, the fragments have been scattered: the i-th fragment is now located at the node with a number k_i! (a factorial of k_i).

To maximize the chance of successful awakening, Ivy decides to place the samples in a node P, so that the total length of paths from each fragment to P is smallest possible. If there are multiple fragments located at the same node, the path from that node to P needs to be counted multiple times.

In the world of zeros and ones, such a requirement is very simple for Ivy. Not longer than a second later, she has already figured out such a node.

But for a mere human like you, is this still possible?

For simplicity, please answer the minimal sum of paths' lengths from every fragment to the emotion samples' assembly node P.

Input

The first line contains an integer n (1 ≤ n ≤ 10^6) — number of fragments of Vanessa's mind.

The second line contains n integers: k_1, k_2, …, k_n (0 ≤ k_i ≤ 5000), denoting the nodes where fragments of Vanessa's mind are located: the i-th fragment is at the node with a number k_i!.

Output

Print a single integer, denoting the minimal sum of path from every fragment to the node with the emotion samples (a.k.a. node P).

As a reminder, if there are multiple fragments at the same node, the distance from that node to P needs to be counted multiple times as well.

Examples

Input


3
2 1 4


Output


5


Input


4
3 1 4 4


Output


6


Input


4
3 1 4 1


Output


6


Input


5
3 1 4 1 5


Output


11

Note

Considering the first 24 nodes of the system, the node network will look as follows (the nodes 1!, 2!, 3!, 4! are drawn bold):

<image>

For the first example, Ivy will place the emotion samples at the node 1. From here:

  * The distance from Vanessa's first fragment to the node 1 is 1. 
  * The distance from Vanessa's second fragment to the node 1 is 0. 
  * The distance from Vanessa's third fragment to the node 1 is 4. 



The total length is 5.

For the second example, the assembly node will be 6. From here:

  * The distance from Vanessa's first fragment to the node 6 is 0. 
  * The distance from Vanessa's second fragment to the node 6 is 2. 
  * The distance from Vanessa's third fragment to the node 6 is 2. 
  * The distance from Vanessa's fourth fragment to the node 6 is again 2. 



The total path length is 6.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
constexpr long long MOD = 1000000007;
constexpr long long INF = 1LL << 60;
constexpr long double EPS = 1e-10;
string operator*(const string& s, int k) {
  if (k == 0) return "";
  string p = (s + s) * (k / 2);
  if (k % 2 == 1) p += s;
  return p;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
struct Edge {
  long long to, cap, rev;
  Edge(long long _to, long long _cap, long long _rev) {
    to = _to;
    cap = _cap;
    rev = _rev;
  }
};
void add_edge(vector<vector<Edge>>& G, long long from, long long to,
              long long cap, bool revFlag, long long revCap) {
  G[from].push_back(Edge(to, cap, (long long)G[to].size()));
  if (revFlag)
    G[to].push_back(Edge(from, revCap, (long long)G[from].size() - 1));
}
long long max_flow_dfs(vector<vector<Edge>>& G, long long v, long long t,
                       long long f, vector<bool>& used) {
  if (v == t) return f;
  used[v] = true;
  for (int i = 0; i < G[v].size(); ++i) {
    Edge& e = G[v][i];
    if (!used[e.to] && e.cap > 0) {
      long long d = max_flow_dfs(G, e.to, t, min(f, e.cap), used);
      if (d > 0) {
        e.cap -= d;
        G[e.to][e.rev].cap += d;
        return d;
      }
    }
  }
  return 0;
}
long long max_flow(vector<vector<Edge>>& G, long long s, long long t) {
  long long flow = 0;
  for (;;) {
    vector<bool> used(G.size());
    for (int(i) = (int)(0); i < (int)(used.size()); i++) used[i] = false;
    long long f = max_flow_dfs(G, s, t, INF, used);
    if (f == 0) {
      return flow;
    }
    flow += f;
  }
}
void BellmanFord(vector<vector<Edge>>& G, long long s, vector<long long>& d,
                 vector<long long>& negative) {
  d.resize(G.size());
  negative.resize(G.size());
  for (int(i) = (int)(0); i < (int)(d.size()); i++) d[i] = INF;
  for (int(i) = (int)(0); i < (int)(d.size()); i++) negative[i] = false;
  d[s] = 0;
  for (int(k) = (int)(0); k < (int)(G.size() - 1); k++) {
    for (int(i) = (int)(0); i < (int)(G.size()); i++) {
      for (int(j) = (int)(0); j < (int)(G[i].size()); j++) {
        if (d[i] != INF && d[G[i][j].to] > d[i] + G[i][j].cap) {
          d[G[i][j].to] = d[i] + G[i][j].cap;
        }
      }
    }
  }
  for (int(k) = (int)(0); k < (int)(G.size() - 1); k++) {
    for (int(i) = (int)(0); i < (int)(G.size()); i++) {
      for (int(j) = (int)(0); j < (int)(G[i].size()); j++) {
        if (d[i] != INF && d[G[i][j].to] > d[i] + G[i][j].cap) {
          d[G[i][j].to] = d[i] + G[i][j].cap;
          negative[G[i][j].to] = true;
        }
        if (negative[i] == true) negative[G[i][j].to] = true;
      }
    }
  }
}
void Dijkstra(vector<vector<Edge>>& G, long long s, vector<long long>& d) {
  d.resize(G.size());
  for (int(i) = (int)(0); i < (int)(d.size()); i++) d[i] = INF;
  d[s] = 0;
  priority_queue<pair<long long, long long>, vector<pair<long long, long long>>,
                 greater<pair<long long, long long>>>
      q;
  q.push(make_pair(0, s));
  while (!q.empty()) {
    pair<long long, long long> a = q.top();
    q.pop();
    if (d[a.second] < a.first) continue;
    for (int(i) = (int)(0); i < (int)(G[a.second].size()); i++) {
      Edge e = G[a.second][i];
      if (d[e.to] > d[a.second] + e.cap) {
        d[e.to] = d[a.second] + e.cap;
        q.push(make_pair(d[e.to], e.to));
      }
    }
  }
}
void WarshallFloyd(vector<vector<Edge>>& G, vector<vector<long long>>& d) {
  d.resize(G.size());
  for (int(i) = (int)(0); i < (int)(d.size()); i++) d[i].resize(G.size());
  for (int(i) = (int)(0); i < (int)(d.size()); i++) {
    for (int(j) = (int)(0); j < (int)(d[i].size()); j++) {
      d[i][j] = ((i != j) ? INF : 0);
    }
  }
  for (int(i) = (int)(0); i < (int)(G.size()); i++) {
    for (int(j) = (int)(0); j < (int)(G[i].size()); j++) {
      chmin(d[i][G[i][j].to], G[i][j].cap);
    }
  }
  for (int(i) = (int)(0); i < (int)(G.size()); i++) {
    for (int(j) = (int)(0); j < (int)(G.size()); j++) {
      for (int(k) = (int)(0); k < (int)(G.size()); k++) {
        chmin(d[j][k], d[j][i] + d[i][k]);
      }
    }
  }
}
bool tsort(vector<vector<Edge>>& graph, vector<int>& order) {
  int n = graph.size(), k = 0;
  vector<long long> in(n);
  for (auto& es : graph)
    for (auto& e : es) in[e.to]++;
  priority_queue<long long, vector<long long>, greater<long long>> que;
  for (int(i) = (int)(0); i < (int)(n); i++)
    if (in[i] == 0) que.push(i);
  while (que.size()) {
    int v = que.top();
    que.pop();
    order.push_back(v);
    for (auto& e : graph[v])
      if (--in[e.to] == 0) que.push(e.to);
  }
  if (order.size() != n)
    return false;
  else
    return true;
}
class lca {
 public:
  const int n = 0;
  const int log2_n = 0;
  std::vector<std::vector<int>> parent;
  std::vector<int> depth;
  lca() {}
  lca(const vector<vector<Edge>>& g, int root)
      : n(g.size()),
        log2_n(log2(n) + 1),
        parent(log2_n, std::vector<int>(n)),
        depth(n) {
    dfs(g, root, -1, 0);
    for (int k = 0; k + 1 < log2_n; k++) {
      for (int v = 0; v < (int)g.size(); v++) {
        if (parent[k][v] < 0)
          parent[k + 1][v] = -1;
        else
          parent[k + 1][v] = parent[k][parent[k][v]];
      }
    }
  }
  void dfs(const vector<vector<Edge>>& g, int v, int p, int d) {
    parent[0][v] = p;
    depth[v] = d;
    for (auto& e : g[v]) {
      if (e.to != p) dfs(g, e.to, v, d + 1);
    }
  }
  int get(int u, int v) {
    if (depth[u] > depth[v]) std::swap(u, v);
    for (int k = 0; k < log2_n; k++) {
      if ((depth[v] - depth[u]) >> k & 1) {
        v = parent[k][v];
      }
    }
    if (u == v) return u;
    for (int k = log2_n - 1; k >= 0; k--) {
      if (parent[k][u] != parent[k][v]) {
        u = parent[k][u];
        v = parent[k][v];
      }
    }
    return parent[0][u];
  }
};
void visit(const vector<vector<Edge>>& g, int v, vector<vector<long long>>& scc,
           stack<long long>& S, vector<long long>& inS, vector<long long>& low,
           vector<long long>& num, int& time) {
  low[v] = num[v] = ++time;
  S.push(v);
  inS[v] = true;
  for (decltype((g[v]).begin()) e = (g[v]).begin(); e != (g[v]).end(); ++e) {
    int w = e->to;
    if (num[w] == 0) {
      visit(g, w, scc, S, inS, low, num, time);
      low[v] = min(low[v], low[w]);
    } else if (inS[w])
      low[v] = min(low[v], num[w]);
  }
  if (low[v] == num[v]) {
    scc.push_back(vector<long long>());
    while (1) {
      int w = S.top();
      S.pop();
      inS[w] = false;
      scc.back().push_back(w);
      if (v == w) break;
    }
  }
}
void stronglyConnectedComponents(const vector<vector<Edge>>& g,
                                 vector<vector<long long>>& scc) {
  const int n = g.size();
  vector<long long> num(n), low(n);
  stack<long long> S;
  vector<long long> inS(n);
  int time = 0;
  for (int(u) = (int)(0); u < (int)(n); u++)
    if (num[u] == 0) visit(g, u, scc, S, inS, low, num, time);
}
class UnionFind {
  vector<int> data;
  long long num;

 public:
  UnionFind(int size) : data(size, -1), num(size) {}
  bool unite(int x, int y) {
    x = root(x);
    y = root(y);
    if (x != y) {
      if (data[y] < data[x]) swap(x, y);
      data[x] += data[y];
      data[y] = x;
    }
    num -= (x != y);
    return x != y;
  }
  bool findSet(int x, int y) { return root(x) == root(y); }
  int root(int x) { return data[x] < 0 ? x : data[x] = root(data[x]); }
  long long size(int x) { return -data[root(x)]; }
  long long numSet() { return num; }
};
class SumSegTree {
 private:
  long long _sum(long long a, long long b, long long k, long long l,
                 long long r) {
    if (r <= a || b <= l) return 0;
    if (a <= l && r <= b)
      return dat[k];
    else {
      long long s1 = _sum(a, b, 2 * k + 1, l, (l + r) / 2);
      long long s2 = _sum(a, b, 2 * k + 2, (l + r) / 2, r);
      return s1 + s2;
    }
  }

 public:
  long long n, height;
  vector<long long> dat;
  SumSegTree(long long _n) {
    n = 1;
    height = 1;
    while (n < _n) {
      n *= 2;
      height++;
    }
    dat = vector<long long>(2 * n - 1, 0);
  }
  void add(long long i, long long x) {
    i += n - 1;
    dat[i] += x;
    while (i > 0) {
      i = (i - 1) / 2;
      dat[i] += x;
    }
  }
  long long sum(long long a, long long b) { return _sum(a, b, 0, 0, n); }
};
class RmqTree {
 private:
  long long _find(long long a, long long b, long long k, long long l,
                  long long r) {
    if (r <= a || b <= l) return INF;
    if (a <= l && r <= b)
      return dat[k];
    else {
      long long s1 = _find(a, b, 2 * k + 1, l, (l + r) / 2);
      long long s2 = _find(a, b, 2 * k + 2, (l + r) / 2, r);
      return min(s1, s2);
    }
  }

 public:
  long long n, height;
  vector<long long> dat;
  RmqTree(long long _n) {
    n = 1;
    height = 1;
    while (n < _n) {
      n *= 2;
      height++;
    }
    dat = vector<long long>(2 * n - 1, INF);
  }
  void update(long long i, long long x) {
    i += n - 1;
    dat[i] = x;
    while (i > 0) {
      i = (i - 1) / 2;
      dat[i] = min(dat[i * 2 + 1], dat[i * 2 + 2]);
    }
  }
  long long find(long long a, long long b) { return _find(a, b, 0, 0, n); }
};
void divisor(long long n, vector<long long>& ret) {
  for (long long i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      ret.push_back(i);
      if (i * i != n) ret.push_back(n / i);
    }
  }
  sort(ret.begin(), ret.end());
}
void divisor_prime(long long n, vector<pair<long long, long long>>& ret) {
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      ret.push_back({i, 0});
      while (n % i == 0) {
        n /= i;
        ret[ret.size() - 1].second++;
      }
    }
  }
  if (n != 1) ret.push_back({n, 1});
}
long long mod_pow(long long x, long long n, long long mod) {
  long long res = 1;
  while (n > 0) {
    if (n & 1) res = res * x % mod;
    x = x * x % mod;
    n >>= 1;
  }
  return res;
}
long long mod_inv(long long x, long long mod) {
  return mod_pow(x, mod - 2, mod);
}
class Combination {
 public:
  vector<long long> fact;
  vector<long long> inv;
  long long mod;
  long long mod_inv(long long x) {
    long long n = mod - 2LL;
    long long res = 1LL;
    while (n > 0) {
      if (n & 1) res = res * x % mod;
      x = x * x % mod;
      n >>= 1;
    }
    return res;
  }
  long long nCr(long long n, long long r) {
    if (n < r) return 0;
    if (n < mod) return ((fact[n] * inv[r] % mod) * inv[n - r]) % mod;
    long long ret = 1;
    while (n || r) {
      long long _n = n % mod, _r = r % mod;
      n /= mod;
      r /= mod;
      (ret *= nCr(_n, _r)) %= mod;
    }
    return ret;
  }
  long long nPr(long long n, long long r) {
    return (fact[n] * inv[n - r]) % mod;
  }
  long long nHr(long long n, long long r) { return nCr(r + n - 1, r); }
  Combination(long long _n, long long _mod) {
    mod = _mod;
    long long n = min(_n + 1, mod);
    fact.resize(n);
    fact[0] = 1;
    for (int(i) = (int)(0); i < (int)(n - 1); i++) {
      fact[i + 1] = (fact[i] * (i + 1LL)) % mod;
    }
    inv.resize(n);
    inv[n - 1] = mod_inv(fact[n - 1]);
    for (int i = n - 1; i > 0; i--) {
      inv[i - 1] = inv[i] * i % mod;
    }
  }
};
long long gcd(long long m, long long n) {
  if (n == 0) return m;
  return gcd(n, m % n);
}
long long lcm(long long m, long long n) { return m / gcd(m, n) * n; }
long long popcount(long long x) {
  x = (x & 0x5555555555555555) + (x >> 1 & 0x5555555555555555);
  x = (x & 0x3333333333333333) + (x >> 2 & 0x3333333333333333);
  x = (x & 0x0F0F0F0F0F0F0F0F) + (x >> 4 & 0x0F0F0F0F0F0F0F0F);
  x = (x & 0x00FF00FF00FF00FF) + (x >> 8 & 0x00FF00FF00FF00FF);
  x = (x & 0x0000FFFF0000FFFF) + (x >> 16 & 0x0000FFFF0000FFFF);
  x = (x & 0x00000000FFFFFFFF) + (x >> 32 & 0x00000000FFFFFFFF);
  return x;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  vector<long long> k(n);
  for (int(i) = (int)(0); i < (int)(n); i++) cin >> k[i];
  long long ans = 0;
  sort((k).begin(), (k).end());
  vector<long long> prime;
  map<long long, long long> p_num;
  for (int(i) = (int)(2); i < (int)(5500); i++) {
    long long flag = 1;
    for (int(j) = (int)(0); j < (int)(prime.size()); j++) {
      if (i % prime[j] == 0) flag = 0;
      if (prime[j] * prime[j] > i) break;
    }
    if (flag) prime.push_back(i), p_num[prime.back()] = prime.size() - 1;
  }
  vector<long long> sum(5001, 0);
  vector<vector<long long>> prime_sum(5001, vector<long long>(prime.size(), 0));
  for (int(i) = (int)(1); i < (int)(5001); i++) {
    vector<pair<long long, long long>> ret;
    divisor_prime(i, ret);
    for (auto itr : ret)
      sum[i] += itr.second, prime_sum[i][p_num[itr.first]] += itr.second;
    sum[i] += sum[i - 1];
    for (int(j) = (int)(0); j < (int)(prime.size()); j++)
      prime_sum[i][j] += prime_sum[i - 1][j];
  }
  for (int(i) = (int)(0); i < (int)(n); i++) {
    ans += sum[k[i]];
  }
  long long l = 0, r = 0;
  long long mx = 0;
  for (int(i) = (int)(0); i < (int)(n); i++) {
    if (k[i] == 1)
      l = i + 1, r = i + 1;
    else if (k[i] >= prime[mx + 1]) {
      if ((r - l) * 2 > n) break;
      l = i, r = i + 1;
      while (k[i] >= prime[mx + 1]) mx++;
    } else
      r++;
  }
  if ((r - l) * 2 <= n) {
    cout << ans << endl;
    return 0;
  }
  long long dif = r - l - (n - r + l);
  for (int i = mx; i >= 0; i--) {
    long long lk = k[l], rk = k[r - 1];
    if ((r - l) * 2 <= n) break;
    dif = r - l - (n - r + l);
    if (prime_sum[lk][i] == prime_sum[rk][i]) {
      ans -= prime_sum[lk][i] * dif;
    } else {
      ans -= prime_sum[lk][i] * dif;
      long long d = prime_sum[lk][i];
      long long lt = l, rt = r;
      while (true) {
        lt++;
        if ((rt - lt) * 2 <= n) break;
        long long dift = rt - lt - (n - rt + lt);
        if (prime_sum[k[lt]][i] > d) {
          ans -= (prime_sum[k[lt]][i] - d) * dift;
          d = prime_sum[k[lt]][i];
        }
      }
      long long cnt = 1;
      for (int(j) = (int)(l + 1); j < (int)(r); j++) {
        if (prime_sum[k[j]][i] == prime_sum[k[j - 1]][i])
          cnt++;
        else {
          if (cnt * 2 > n) {
            r = j;
            l = r - cnt;
            break;
          } else
            cnt = 1;
        }
      }
      if (cnt * 2 > n)
        l = r - cnt;
      else
        break;
    }
  }
  cout << ans << "\n";
  return 0;
}